# Thread: Pappus's Theorem and Some Integrals

1. ## Pappus's Theorem and Some Integrals

Hey guys, I was wondering if it would be possible for someone to check my work on a few calculus problems and, if they're wrong, to tell me what I'm doing wrong. Thank you very much in advance.

Problem 1. R is the region bounded by y = sinx and the lines x = 0 and x = $\frac{\pi}{2}$.

a. Calculate the center of mass of the region R.

My Solution

$\overline{x} = \frac{\int^{\frac{\pi}{2}}_{0}xsinx}{\int^{\frac{\ pi}{2}}_{0}sinx}$ dx

u = x, du = dx, dv = sinx dx, v = cosx

-xcosx + $\int$cosx dx = -xcosx + sinx

$\frac{[-xcosx + sinx]^{\frac{\pi}{2}}_{0}}{[-cosx]^{\frac{\pi}{2}}_{0}}$ = $\frac{1 - 0}{0 + 1}$ = 1

$\overline{y} = \frac{\frac{1}{2}\int^{\frac{\pi}{2}}_{0}\sin^2(x) }{\int^{\frac{\pi}{2}}_{0}sinx}$ dx = $\frac{\frac{1}{2}\int^{\frac{\pi}{2}}_{0}\frac{1 - cos(2x)}{2}}{[-cosx]^{\frac{\pi}{2}}_{0}}$ dx = $\frac{\frac{1}{4}\int^{\frac{\pi}{2}}_{0}1 - cos(2x)}{0 + 1}$ dx = $\frac{\frac{1}{4}[x - \frac{1}{2}sin(2x)]^{\frac{\pi}{2}}_{0}}{1}$ = $\frac{\frac{1}{4}(\frac{\pi}{2}) - 0}{1}$ = $\frac{\frac{\pi}{8}}{1}$ = $\frac{\pi}{8}$

b. Use Pappus's Theorem to calculate the volume of the solid of revolution if R is revolved about the line x = 5.

My Solution

Area of region R = 1 (from problem 1a)

V = $2\pi(5 - \frac{\pi}{2})(1)$ = $2\pi(\frac{10 - \pi}{2})$ = $10\pi - \pi^2$

c. Calculate the volume of the solid if R is revolved about the line y = -4.

My Solution

Area of region R = 1 (from problem 1a)

V = $2\pi(\frac{\pi}{8} + 4)(1)$ = $2\pi(\frac{\pi + 32}{8})$ = $\frac{\pi^2 + 32\pi}{4}$

Problem 2.
Calculate each antiderivative.

a. $\int\frac{1}{\sqrt{1-x^2}}$ dx

My Solution

x = $\sin\theta$, dx = $\cos\theta$ d $\theta$, $\theta$ = $\arcsin{x}$

= $\int\frac{1}{\sqrt{1 - \sin^2(\theta)}}\cos(\theta)$ d $\theta$ = $\int\frac{1}{\sqrt{\cos^2(\theta)}}\cos(\theta)$ d $\theta$ = $\int$d $\theta$ = $\theta + C$ = $\arcsin{x} + C$

b. $\int\frac{x}{\sqrt{1 - x^2}}$ dx

My Solution

u = 1 - $x^2$, du = -2x dx, dx = $\frac{-1}{2x}$ du

= $\int\frac{x}{u^\frac{1}{2}}(\frac{-1}{2x})$ du = $\frac{-1}{2}\int$ $\frac{1}{u^\frac{1}{2}}$ du = $\frac{-1}{2}\int$ $u^\frac{-1}{2}$ du = $-\sqrt{u} + C$ = $-\sqrt{1 - x^2} + C$

c. $\int\frac{x^2}{\sqrt{1-x^2}}$ dx

My Solution

$x = sin\theta$, $dx = cos\theta$ $d\theta$, $\theta = \arcsin{x}$

= $\int\frac{\sin^2(\theta)}{\sqrt{1 - \sin^2(\theta)}}cos\theta$ $d\theta$ = $\int\frac{\sin^2(\theta)}{\sqrt{\cos^2(\theta)}}co s(\theta)$ d $\theta$ = $\int\frac{\sin^2(\theta)}{cos(\theta)}cos(\theta)$ d $\theta$ = $\int\sin^2(\theta)$ $d\theta$ = $\int\frac{1 + cos(2\theta)}{2}$ $d\theta$ = $\frac{1}{2}\int$ $1 + cos(2\theta)$ $d\theta$ = $\frac{1}{2}(\theta + \frac{1}{2}\sin(2\theta))$ = $\frac{1}{2}\arcsin{x} + \frac{1}{4}\sin(2x) + C$

d. $\int\frac{x^3}{\sqrt{1 - x^2}}$ dx

My Solution

$x = \sin(\theta)$, $dx = \cos(\theta)$ $d\theta$, $\theta = \arcsin{x}$

= $\int\frac{\sin^3(\theta)}{\sqrt{1 - \sin^2(\theta)}}\cos(\theta)$ $d\theta$ = $\int\frac{\sin^3(\theta)}{\sqrt{\cos^2(\theta)}}\c os(\theta)$ $d\theta$ = $\int\frac{\sin^3(\theta)}{cos(\theta)}\cos(\theta)$ $d\theta$ = $\int\sin^3(\theta)$ $d\theta$ = $\int\sin(\theta)(\frac{1 + cos(2\theta)}{2})$ $d\theta$ = $\frac{1}{2}\int\sin(\theta)(1 + cos(2\theta))$ $d\theta$ = $\frac{1}{2}\int\sin(\theta) + \sin(\theta)\cos(2\theta) d\theta$ = $\frac{1}{4}[-\cos(\theta) + \cos^2(2\theta)] + C$ = $\frac{1}{4}[\sqrt{1 - x^2} - 2(1 - x^2)] + C$

Once again, thank you very much in advance.

2. Originally Posted by mturner07
c. $\int\frac{x^2}{\sqrt{1-x^2}}$ dx

My Solution

$x = sin\theta$, $dx = cos\theta$ $d\theta$, $\theta = \arcsin{x}$

= $\int\frac{\sin^2(\theta)}{\sqrt{1 - \sin^2(\theta)}}cos\theta$ $d\theta$ = $\int\frac{\sin^2(\theta)}{\sqrt{\cos^2(\theta)}}co s(\theta)$ d $\theta$ = $\int\frac{\sin^2(\theta)}{cos(\theta)}cos(\theta)$ d $\theta$ = $\int\sin^2(\theta)$ $d\theta$ = $\int\frac{1 {\color{red}+} cos(2\theta)}{2}$ $d\theta$ = $\frac{1}{2}\int$ $1 {\color{red}+} cos(2\theta)$ $d\theta$ = $\frac{1}{2}(\theta {\color{red}+} \frac{1}{2}\sin(2\theta))$ = $\frac{1}{2}\arcsin{x} {\color{red}+} \frac{1}{4}\sin({2{\color{red}x}}) + C$
And really I like your way to post all work.
I wonder if every asker can do that!
Note: I did not check the other problems, It still maybe correct or wrong.

3. Originally Posted by mturner07
d. $\int\frac{x^3}{\sqrt{1 - x^2}}$ dx

My Solution

$x = \sin(\theta)$, $dx = \cos(\theta)$ $d\theta$, $\theta = \arcsin{x}$

= $\int\frac{\sin^3(\theta)}{\sqrt{1 - \sin^2(\theta)}}\cos(\theta)$ $d\theta$ = $\int\frac{\sin^3(\theta)}{\sqrt{\cos^2(\theta)}}\c os(\theta)$ $d\theta$ = $\int\frac{\sin^3(\theta)}{cos(\theta)}\cos(\theta)$ $d\theta$ = $\int\sin^3(\theta) d\theta$
You just make it complicated from here.
Notice that:
$\int\sin^3(\theta) d\theta=\int sin^2(\theta) sin(\theta) d\theta=\int (1-cos^2(\theta)) sin(\theta) d\theta$
Use $u=cos(\theta)$.

4. Originally Posted by General
You just make it complicated from here.
Notice that:
$\int\sin^3(\theta) d\theta=\int sin^2(\theta) sin(\theta) d\theta=\int (1-cos^2(\theta)) sin(\theta) d\theta$
Use $u=cos(\theta)$.
Note for (d) that you could also do the following:

Let $u=1-x^2\implies x^2=1-u$. This implies that $\,du=-2x\,dx$

The integral would then become $-\tfrac{1}{2}\int\frac{1-u}{\sqrt{u}}\,du=\tfrac{1}{2}\int u^{1/2}-u^{-1/2}\,du$

5. Thanks for the replies. Yeah it takes awhile to type up, but if people can't see my work they can't direct me to where I went wrong lol.

Now for 2c. I get $\frac{1}{2}\arcsin{x} - \frac{x}{4}\sqrt{1 - x^2} + C$ and for problem 2d. I get $-\sqrt{1 - x^2} + \frac{(1 - x^2)\sqrt{1 - x^2}}{3} + C$. Are these solutions, along with the other problems I posted correct? Thanks in advance.

6. Anyone?